4 02/04 F

1. 1st Ed: p103, q44, 2nd Ed: p123, q44. Do it with Stokes.

2. 1st Ed: p104, q62, 2nd Ed: p124, q62. (20 points) Do both directly and using the divergence theorem. Make sure to include the base of the cone. Use the Cartesian expression for $\vec{n} {\rm d}{S}$ to formulate the surface integral, then switch to polar to do it.

3. 1st Ed: p132, q50, 2nd Ed: p154, q50 MODIFIED. Given
   
   $$\vec v = \frac{(-y,x)}{x^2+y^2}$$
   
   
   evaluate $\nabla\times\vec{v}$. Also evaluate, presumably using polar coordinates,
   
   $$\oint_{\rm I} \vec v \cdot {\rm d}\vec r \qquad \oint_{\rm II} \vec v \cdot {\rm d}\vec r$$
   
   
   where path I is the semi circle of radius $r$ going clockwise from $(r,0)$ to $(-r,0)$, and path II is the semi circle of radius $r$ going counter-clockwise from $(r,0)$ to $(-r,0)$. Explain why the integral over II minus the integral over I is the integral over the closed circle. Explain why Stokes implies that the closed contour integral should be the integral of the $z$-compont of $\nabla\times\vec{v}$ over the inside of the circle. Then explain why you would then normally expect the contour integral to be zero. That means that the two integrals I and II should be equal, but they are not. Explain what the problem is.

   Do you expect integrals over closed contours of different radii to be equal? Why? Are they equal?

   Now assume that you allow singular functions to be OK, like Heaviside step functions and Dirac delta functions say. Then figure out in what part of the interior of the circle, $\int\!\int\nabla\times\vec{v}\cdot{\hat k} {\rm d}{x}{\rm d}{y}$ is not zero. So how would you describe $\nabla\times\vec{v}$ for this vector field in terms of singular functions?

4. 1st Ed: p133, q56, 2nd Ed: p155, q56.

5. Read through subsection 10.4 of QMFE and write a half-page summary.

6. Derive $\vec n {\rm d}S$ in terms of ${\rm d}\theta$ and ${\rm d}\phi$, where $(r,\theta,\phi)$ are spherical coordinates, assuming that the surface is given as $r=f(\theta,\phi)$. Use that:
   
   $$\vec r = r \hat\imath_r \quad \frac{\partial \hat\imat... ...al \hat\imath_r}{\partial\phi} = \sin\theta \hat\imath_\phi$$
   
   
   as will be derived in class. Next generalize the result to the case that the surface is given by the implicit expression $F(r,\theta,\phi)=0$. One way to do so is to find $\partial r/\partial\theta$ and $\partial r/\partial\phi$ from the total differential
   
   $$\frac{\partial F}{\partial r} {\rm d}r + \frac{\partial F}... ...m d}\theta + \frac{\partial F}{\partial\phi} {\rm d}\phi = 0$$
   
   

   Now clean up your result to something similar to the vector expression that you have in Cartesian coordinates,
   

   $$\vec n {\rm d}S = \frac{\nabla F}{F_z} {\rm d}x {\rm d}y$$
   
   
   as derived in class. You may want to look up the gradient in spherical coordinates, and the infinitesimal volume element.

7. 1st Ed: p160, q38, 2nd Ed: p183, q38. Simplify as much as possible.